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Stolb23 [73]
3 years ago
13

Look who’s back, back again. Can you help me?

Mathematics
2 answers:
Maksim231197 [3]3 years ago
7 0
Y=2x+8
Brainliest answer?

Neporo4naja [7]3 years ago
4 0

desmos is good for graphing problems

Step-by-step explanation:

You might be interested in
Write the equation of each line using the given information.
MrRissso [65]

Answer:

See explanation.

Step-by-step explanation:

We need a point and a slope to have a point slope equation.

y-y_{1}=m(x-x_{1})

We need a slope and y-intercept to write in slope intercept form.

y=mx+b

A) y-1=\frac{1}{2}(x+4) or y=\frac{1}{2}x+3

B) y+1=-1(x-2)

C) y-1=0(x-1)

D) y=-3x+8

4 0
3 years ago
Ian's monthly allowance is $21. In January he starts saving for a birthday gift in June. Each month he saves of his allowance. T
Dmitrij [34]

Answer:

<h2>Ian will have enough money to buy the gift</h2>

Step-by-step explanation:

Step one:

given

monthly allowance= $21

from January to June= 6 months

total savings = 21*6= $126

cost of gift = $110

Step two:

Ian will have enough money to buy the gift

since he saved $126 which is greater than the cost of the gift of $110

he would have a balance of $16

4 0
3 years ago
Many elementary school students in a school district currently have ear infections. A random sample of children in two different
Marta_Voda [28]

Answer:

Step-by-step explanation:

The summary of the given data includes;

sample size for the first school n_1 = 42

sample size for the second school n_2  = 34

so 16 out of 42 i.e x_1 = 16 and 18 out of 34 i.e x_2 = 18 have ear infection.

the proportion of students with ear infection Is as follows:

\hat p_1 = \dfrac{16}{42} = 0.38095

\hat p_2 = \dfrac{18}{34}  =  0.5294

Since this is a two tailed test , the null and the alternative hypothesis can be computed as :

H_0 :p_1 -p_2 = 0 \\ \\ H_1 : p_1 - p_2 \neq 0

level of significance ∝ = 0.05,

Using the table of standard normal distribution, the value of z that corresponds to the two-tailed probability 0.05 is 1.96. Thus, we will reject the null hypothesis if the value of the test statistics is less than -1.96 or more than 1.96.

The test statistics for the difference in proportion can be achieved by using a pooled sample proportion.

\bar p = \dfrac{x_1 +x_2}{n_1 +n_2}

\bar p = \dfrac{16 +18}{42 +34}

\bar p = \dfrac{34}{76}

\bar p = 0.447368

\bar p + \bar  q = 1 \\ \\ \bar q = 1 -\bar  p \\  \\\bar q = 1 - 0.447368 \\ \\\bar q = 0.552632

The pooled standard error can be computed by using the formula:

S.E = \sqrt{ \dfrac{ \bar p \bar q}{ n_1} +  \dfrac{\bar p \bar p}{n_2} }

S.E = \sqrt{ \dfrac{  0.447368 *  0.552632}{ 42} +  \dfrac{ 0.447368 *  0.447368}{34} }

S.E = \sqrt{ \dfrac{  0.2472298726}{ 42} +  \dfrac{ 0.2001381274}{34} }

S.E = \sqrt{ 0.01177284105}

S.E = 0.1085

The test statistics is ;

z = \dfrac{\hat p_1 - \hat p_2}{S.E}

z = \dfrac{0.38095- 0.5294}{0.1085}

z = \dfrac{-0.14845}{0.1085}

z = - 1.368

Decision Rule: Since the test statistics is greater than the rejection region - 1.96 , we fail to reject the null hypothesis.

Conclusion: There is insufficient evidence to support the claim that a difference exists between the proportions of students who have ear infections at the two schools

5 0
3 years ago
Evaluate the equation. 193b = 212.3
trasher [3.6K]

Answer:

b= 1.1

Step-by-step explanation:

193b= 212.3

divide both sides by 193 to isolate b

212.3÷193=1.1

3 0
2 years ago
(-9,-2) and (1,3) <br> is y-3=1/2(x-1)
Simora [160]

Answer:

(1,3) is a solution.

Step-by-step explanation:

7 0
3 years ago
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